Có: \(\sqrt{a}\sqrt{b}=\sqrt{âb}\)

\(\Rightarrow\sqrt{20}=\sqrt{4}\sqrt{5}=2\sqrt{5}\)

\(\sqrt{20}=\sqrt{2^2.5}=\sqrt{5}.\sqrt{2^2}=2\sqrt{5}\)

\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11+\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)

\(=\sqrt{7-2\sqrt{6}}-\dfrac{5\left(\sqrt{2}-1\right)}{\sqrt{5}\left(\sqrt{2}-1\right)}+\left|11+2\sqrt{30}\right|\sqrt{11-2\sqrt{30}}\)

\(=\sqrt{1^2-2\sqrt{6}\cdot1+\left(\sqrt{6}\right)^2}-\dfrac{\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{5}\cdot\sqrt{6}+\left(\sqrt{5}\right)^2}\)

\(=\sqrt{\left(1-\sqrt{6}\right)^2}-\sqrt{5}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\)

\(=\left|1-\sqrt{6}\right|-\sqrt{5}+\left(11+2\sqrt{30}\right)\left|\sqrt{6}-\sqrt{5}\right|\)

\(=-1+6-\sqrt{5}+\left(\sqrt{6}+\sqrt{5}\right)^2\left(\sqrt{6}-\sqrt{5}\right)\)

\(=\sqrt{6}-1-\sqrt{5}+\left[\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2\right]\left(\sqrt{6}+\sqrt{5}\right)\)

\(=\sqrt{6}-1-\sqrt{5}+\left(6-5\right)\left(\sqrt{6}+\sqrt{5}\right)\)

\(=\sqrt{6}-1-\sqrt{5}+\sqrt{6}+\sqrt{5}\)

\(=2\sqrt{6}-1\)

\(=\sqrt{6+1-2\sqrt{6}}-\dfrac{\sqrt{5}\left(\sqrt{10}-\sqrt{5}\right)}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+\sqrt{120}\right)^2}\\ =\sqrt{\left(\sqrt{6}-\sqrt{1}\right)^2}-\sqrt{5}+\sqrt{\left(11^2-120\right)\left(11+2\sqrt{30}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{1\left(6+5+2\sqrt{6\cdot5}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}-\sqrt{1}\)

Bài 1 : 

\(a.\sqrt{x^2-1}\)

\(ĐK:\)

\(x^2-1\ge0\)

\(\Leftrightarrow x^2\ge1\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)

Bài 2 : 

\(2\cdot\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{48}-5\sqrt{50}\)

\(=2\cdot\left|\sqrt{2}-3\right|+4\sqrt{3}-25\sqrt{2}\)

\(=-2\cdot\left(\sqrt{2}-3\right)+4\sqrt{3}-25\sqrt{2}\)

\(=-2\sqrt{2}-6+4\sqrt{3}-25\sqrt{2}\)

\(=-27\sqrt{2}-6+4\sqrt{3}\)

\(\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1^2}\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=|\sqrt{5}-1|\)

\(=\sqrt{5}-1\)

_Vi hạ_

=\(\left(3\sqrt{3}-3\sqrt{3}+2\sqrt{6}\right):3\sqrt{3}\)
\(=1-\dfrac{\sqrt{6}}{2}+\dfrac{2\sqrt{2}}{3}\)
=\(\dfrac{6}{6}-\dfrac{3\sqrt{6}}{6}+\dfrac{4\sqrt{2}}{6}\)
=\(\dfrac{6+\sqrt{6}}{6}\)

Đặt \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)

\(\Leftrightarrow A^3=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\cdot\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(\Leftrightarrow A^3=4+3\cdot\left(-1\right)\cdot A\)

\(\Leftrightarrow A^3=4-3A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A^2+A^2-A+4A-4=0\)

\(\Leftrightarrow A^2\left(A-1\right)+A\left(A-1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)

\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)

\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)

\(=3\sqrt{2}\)

\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)

\(=\dfrac{3-\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\dfrac{3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

\(=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}\)

\(=\dfrac{3}{2}\)

\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)

\(A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)

\(B=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}=\dfrac{6}{4}=\dfrac{3}{2}\)

\(C=\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)

\(C=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)

\(C=2-\sqrt{3}+3+\sqrt{3}=5\)

\(\dfrac{\sqrt{AC^2+\left(v_n.\dfrac{t_1}{2}\right)^2}}{v}=1\)

\(\Rightarrow AC^2+v^2_n.\dfrac{t^2_1}{4}=v^2\)

\(\Rightarrow v_n^2.\dfrac{t^2_1}{4}=v^2-AC^2\)

\(\Rightarrow t_1^2=\dfrac{4\left(v^2-AC^2\right)}{v_n^2}\Rightarrow t_1=\dfrac{2\sqrt{v^2-AC^2}}{v_n}\)

Bạn ơi, muốn chuyển đổi như vậy thì phải có dấu " = " chứ?

Căng

\(\dfrac{1}{5}\sqrt[]{25x+50}-5\sqrt[]{x+2}+\sqrt[]{9x+18}+9=0\)

\(\Leftrightarrow\dfrac{1}{5}\sqrt[]{25\left(x+2\right)}-5\sqrt[]{x+2}+\sqrt[]{9\left(x+2\right)}+9=0\)

\(\Leftrightarrow\dfrac{1}{5}.5\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}\left(1-5+3\right)+9=0\)

\(\Leftrightarrow-\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}=9\)

\(\Leftrightarrow x+2=81\)

\(\Leftrightarrow x=79\)